## Friday, August 22, 2008

### AP Physics & Degree Entrance Kinematics- Questions on One Dimensional Motion

The following questions numbered 1, 2 and 3 are based on the velocity-time graph of a particle in one dimensional motion shown in the adjoining figure.
(1) The displacement of the particle during the first second of its motion is nearly
(a) 0.25 m
(b) 0.5 m
(c) 1 m
(d) 1.5 m
(e) 2 m
The area under the velocity-time curve gives the displacement. The portion of the graph for the interval from zero to 1 second is straight and the area under the curve is triangular and is equal to (½)×1×4 = 2 m.
(2) The average accelerations during the 3rd second and 7th second respectively are nearly
(a) 0.5 ms–2 and 2 ms–2
(b) 1 ms–2 and –2 ms–2
(c) 1 ms–2 and 2 ms–2
(d) 1 ms–2 and –2 ms–2
(e) zero and –2 ms–2
The 3rd second is the interval from 2 seconds to 3 seconds and during this time the velocity increases from 7 ms–1 to 8 ms–1. The acceleration is therefore (8 – 7)/1 = 1 ms–2.
The 7th second is the interval from 6 seconds to 7 seconds and during this time the velocity decreases from 8 ms–1 to 6 ms–1.
The acceleration is therefore (6 – 8)/1 = – 2 ms–2. [Option (b)]
(3) Which one among the following acceleration–time graphs most closely represents the motion of the particle?
The acceleration is positive and uniform initially. Afterwards the acceleration becomes zero (since the velocity remains constant) for some time and then becomes negative (since the velocity goes on decreasing) and uniform. The curve shown in (b) therefore represents the motion of the particle.
Let us leave the velocity time graph here and consider a couple of different questions:
(4) Two boys running at uniform speeds v1 and v2 respectively along a straight line path in opposite directions get 9 m closer each second. While running along the same direction with their speeds reduced by 50%, they get 0.5m closer each second. The speeds v1 and v2 are respectively
(a) 6 ms–1 and 3 ms–1
(b) 5 ms–1 and 4 ms–1
(c) 5 ms–1 and 4.5 ms–1
(d) 5 ms–1 and 3.5 ms–1
(e) 5.5 ms–1 and 3.5 ms–1
This is a simple question involving relative velocity. Wile running along opposite directions, we have
v1+ v2 = 9
While running along the same direction, we have
v1/2 v2/2 = 0.5, from which v1v2 = 1
Solving the above equations, we obtain v1 = 5 ms–1 and v2 = 4 ms–1
(5) The rear end of a train running on a straight track with uniform acceleration has velocities 6 ms–1 and 10 ms–1 respectively when passing points A and B in its path. The velocity of the rear end midway between these points is approximately
(a) 7 ms–1
(b) 7.5 ms–1
(c) 8ms–1
(d) 8.2 ms–1
(e) 8.4 ms–1
We have v2 = u2 + 2as where u and v are the initial ans final velocities respectively, a is the acceleration and s is the displacement.
If the distance between A and B is s, we have
102 = 62 + 2as from which 2as = 64
If v1 is the velocity midway between A and B we have
v12 = 62 + 2a(s/2) = 62 + 32 = 68
Therefore, v1 = 8.2 ms–1 nearly.
You will find more questions (with solution) on one dimensional motion and other sections at AP Physics Resources

## Saturday, August 09, 2008

### Multiple Choice Questions (MCQ) Involving Rotation

Questions on rotational motion have been discussed on many occasions on this site. You can access them by clicking on the label ‘rotation’ below this post. Now, see the following questions involving rotation.

(1) A light inextensible string is wound round a wheel of moment of inertia I and radius R. A mass m is attached to its hanging end as shown. The wheel and the mass are initially at rest. Assume that there are no frictional forces opposing the motion. When the mass is released, it moves down and rotates the wheel. When the mass has fallen down through a distance h, what is the angular velocity of the wheel?

(b) [2mgh/ (I + 2m)]1/2

(c) [2mgh/ (I + 2mR2)]1/2

(d) [2mgh/ (I + mR2)]1/2

(e) [2mgh/ (I + m)]1/2

The mass loses gravitational potential energy and gains kinetic energy. The wheel also gains kinetic energy so that we have

mgh = ½ mv2 + ½ 2 where v is the velocity of the mass after falling through the distance h and ω is the angular velocity of the wheel.

Since v =ωR the above equation becomes

mgh = ½ 2R 2 + ½ 2

This yields ω = [2mgh/ (I + mR2)]1/2

(2) A simple pendulum has a spherical bob of mass 200 g. The string of the pendulum has negligible mass and it can withstand a maximum tension of 22 N. By holding the string in hand a student whirls the bob in a vertical circle of radius 1 m. The maximum possible angular velocity the bob can have throughout its motion (in radian per second) will be nearly

(a) 2

(b) 4

(c) 6

(d) 8

The tension in the string will be maximum when the bob is at the bottom point of the vertical circle. This can be seen by writing the expression for the resultant force on the bob and equating it to the centripetal force which makes the bob move along the circle:

T mg = mrω2, where m is the mass of the bob, T is the tension in the string, r is the radius of the circular path and ω is the angular velocity of revolution of the bob.

This gives ω = [(T mg)/mr]1/2

= [(22 – 0.2×10)/(0.2×1)]1/2 = 10 radian/sec

We have substituted the maximum tension the string can withstand and hence the above angular velocity is the maximum value possible.

You will find some useful posts on rotational motion at AP Physics Resources: AP Physics B and C– Multiple Choice Questions on Circular Motion and Rotation